Compute the length of the segment tangent from the origin to the circle that passes through the points $(3,4),$ $(6,8),$ and $(5,13).$
Solution: Let $O = (0,0),$ $A = (3,4),$ $B = (6,8),$ and $C = (5,13).$  Let $T$ be a point on the circumcircle of triangle $ABC,$ so that $\overline{OT}$ is tangent to the circumcircle.  Note that $O,$ $A,$ and $B$ are collinear.

[asy]
unitsize(0.4 cm);

pair A, B, C, O, T;

A = (3,4);
B = (6,8);
C = (5,13);
O = circumcenter(A,B,C);
T = intersectionpoints(Circle(O/2,abs(O)/2),circumcircle(A,B,C))[1];

draw(circumcircle(A,B,C));
draw((0,0)--(6,8));
draw((0,0)--T);
draw((-10,0)--(10,0));
draw((0,-2)--(0,18));

label("$O = (0,0)$", (0,0), SW);

dot("$A = (3,4)$", A, SE);
dot("$B = (6,8)$", B, E);
dot("$C = (5,13)$", C, NE);
dot("$T$", T, SW);
[/asy]

Then by power of a point, $OT^2 = OA \cdot OB = 5 \cdot 10 = 50,$ so $OT = \sqrt{50} = \boxed{5 \sqrt{2}}.$